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sergiy2304 [10]

3 years ago

15

How is the basic structure of diamond related to its bulk properties?

1 answer:

KATRIN_1 [288]3 years ago

4 0

Diamonds structure is defined by series of multiple covalent bonds. Given that it is made of a very strong bond, this makes some of its bulk properties relatable to this. Example its density which is larger than most of the crystalline. Another is its intensive property of its temperature, given that the diamond is hard to break down, it also has high melting point.

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A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

2 years ago

Martina has a sample of an unknown substance. She measures the substance. It’s mass is 13.5 grams, and it’s volume is 5 cm3 .Whi

Dominik [7]

Cm^3 is same as mL

13.5 g / 5 mL = 2.7 g/mL
look up densities of metals
aluminum has a density of 2.7 g/mL

6 0

2 years ago

One gallon of paint (volume = 3.79 10-3 m3) covers an area of 17.8 m2. What is the thickness of the fresh paint on the wall?

AURORKA [14]

Answer:

0.2129 mm

Explanation:

We have given volume of the paint = 1 gallon $=3.79\times 10^{-3}m^3$

Area that covers the paint $=17.8 m^2$

We have to find the thickness of the fresh paint

So $thickness=\frac{volume}{area}=\frac{3.79\times 10^{-3}}{17.8}=0.2129\times 10^{-3}m=0.2129mm$

So the thickness of fresh paint on the wall is 0.2129 mm

6 0

3 years ago

Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i

Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

$E=-E_0 \frac{1}{n^2}$

where

$E_0 = 13.6 eV$ is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

$\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})$

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

$\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV$

While the difference between the levels n=2 and n=3 is

$\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV$

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0

3 years ago

Which of the following is not a latent heat gain source

Setler [38]

The pathway for you to be able is in your room you need

6 0

1 year ago